Find the average velocity in the pipe. (See drawing in next comment)

$$v_z=0.1(1-\frac{r^2}{16})$$

My book gives:

$$v_{avg}=\frac{\int\limits \int\limits_A \rho vdA}{\int\limits \int\limits_A \rho dA}$$
Which should simplify given rho is constant, and the same:
$$v_{avg}=\frac{\int\limits \int\limits_A vdA}{\int\limits \int\limits_A dA}$$
It seems strange to perform double integration since the velocity is a function of a single variable.

So I assume I can rewrite it as:
$$v_{avg}=\frac{\int\limits_L vdL}{\int\limits_L dL}$$
Where$$4 \le L \le -4$$

Is this valid?

Question

Find the average velocity in the pipe. (See drawing in next comment)

$$v_z=0.1(1-\frac{r^2}{16})$$

My book gives:

$$v_{avg}=\frac{\int\limits \int\limits_A \rho vdA}{\int\limits \int\limits_A \rho dA}$$
Which should simplify given rho is constant, and the same:
$$v_{avg}=\frac{\int\limits \int\limits_A  vdA}{\int\limits \int\limits_A dA}$$
It seems strange to perform double integration since the velocity is a function of a single variable.

So I assume I can rewrite it as:
$$v_{avg}=\frac{\int\limits_L  vdL}{\int\limits_L dL}$$
Where$$4 \le L \le -4$$

Is this valid?

anonymous · Answer

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